Math worksheet with algebra problems and answers related to quadratic equations, themed around eggs.

To solve the problems in the image, we need to factorize the given quadratic expressions. Let's go through each problem step by step.

Problem 1: \(3y^2 + 16y + 5\)

We need to factorize \(3y^2 + 16y + 5\).

1. Identify coefficients: \(a = 3\), \(b = 16\), \(c = 5\).
2. Find two numbers whose product is \(a \cdot c = 3 \cdot 5 = 15\) and whose sum is \(b = 16\). The numbers are \(15\) and \(1\) because \(15 \cdot 1 = 15\) and \(15 + 1 = 16\).
3. Rewrite the middle term: \(3y^2 + 15y + y + 5\).
4. Group and factor: \((3y^2 + 15y) + (y + 5) = 3y(y + 5) + 1(y + 5)\).
5. Factor out the common term: \((3y + 1)(y + 5)\).

So, the factorization is \((3y + 1)(y + 5)\).

Problem 2: \(6z^2 - 11z - 2\)

We need to factorize \(6z^2 - 11z - 2\).

1. Identify coefficients: \(a = 6\), \(b = -11\), \(c = -2\).
2. Find two numbers whose product is \(a \cdot c = 6 \cdot (-2) = -12\) and whose sum is \(b = -11\). The numbers are \(-12\) and \(1\) because \(-12 \cdot 1 = -12\) and \(-12 + 1 = -11\).
3. Rewrite the middle term: \(6z^2 - 12z + z - 2\).
4. Group and factor: \((6z^2 - 12z) + (z - 2) = 6z(z - 2) + 1(z - 2)\).
5. Factor out the common term: \((6z + 1)(z - 2)\).

So, the factorization is \((6z + 1)(z - 2)\).

Problem 3: \(5w^2 - 9w + 2\)

We need to factorize \(5w^2 - 9w + 2\).

1. Identify coefficients: \(a = 5\), \(b = -9\), \(c = 2\).
2. Find two numbers whose product is \(a \cdot c = 5 \cdot 2 = 10\) and whose sum is \(b = -9\). The numbers are \(-5\) and \(-4\) because \(-5 \cdot (-4) = 10\) and \(-5 + (-4) = -9\).
3. Rewrite the middle term: \(5w^2 - 5w - 4w + 2\).
4. Group and factor: \((5w^2 - 5w) + (-4w + 2) = 5w(w - 1) - 2(2w - 1)\).
5. Factor out the common term: \((5w - 2)(w - 1)\).

So, the factorization is \((5w - 2)(w - 1)\).

Problem 4: \(4x^2 + 2x + 35\)

This quadratic expression does not factorize into real numbers because the discriminant (\(b^2 - 4ac\)) is negative. Therefore, it cannot be factored over the real numbers.

Problem 5: \(6y - 11y - 10\)

This is not a quadratic expression. It simplifies to \(-5y - 10\), which can be factored as \(-5(y + 2)\).

Problem 6: \(3z - 3z + 44\)

This simplifies to \(44\), which is a constant and cannot be factored further.

Problem 7: \(10m + 17m + 6\)

This simplifies to \(27m + 6\), which can be factored as \(3(9m + 2)\).

Problem 8: \(1y^2 - 15y + 4\)

We need to factorize \(y^2 - 15y + 4\).

1. Identify coefficients: \(a = 1\), \(b = -15\), \(c = 4\).
2. Find two numbers whose product is \(a \cdot c = 1 \cdot 4 = 4\) and whose sum is \(b = -15\). The numbers are \(-1\) and \(-14\) because \(-1 \cdot (-14) = 4\) and \(-1 + (-14) = -15\).
3. Rewrite the middle term: \(y^2 - y - 14y + 4\).
4. Group and factor: \((y^2 - y) + (-14y + 4) = y(y - 1) - 4(3y - 1)\).
5. Factor out the common term: \((y - 1)(y - 4)\).

So, the factorization is \((y - 1)(y - 4)\).

Problem 9: \(4z + 32z + 63\)

This simplifies to \(36z + 63\), which can be factored as \(9(4z + 7)\).

Problem 10: \(6r^2 + 7r + 10\)

This quadratic expression does not factorize into real numbers because the discriminant (\(b^2 - 4ac\)) is negative. Therefore, it cannot be factored over the real numbers.

Problem 11: \(8d^2 - 2d - 3\)

We need to factorize \(8d^2 - 2d - 3\).

1. Identify coefficients: \(a = 8\), \(b = -2\), \(c = -3\).
2. Find two numbers whose product is \(a \cdot c = 8 \cdot (-3) = -24\) and whose sum is \(b = -2\). The numbers are \(-6\) and \(4\) because \(-6 \cdot 4 = -24\) and \(-6 + 4 = -2\).
3. Rewrite the middle term: \(8d^2 - 6d + 4d - 3\).
4. Group and factor: \((8d^2 - 6d) + (4d - 3) = 2d(4d - 3) + 1(4d - 3)\).
5. Factor out the common term: \((2d + 1)(4d - 3)\).

So, the factorization is \((2d + 1)(4d - 3)\).

Problem 12: \(2k^2 + 19k - 10\)

We need to factorize \(2k^2 + 19k - 10\).

1. Identify coefficients: \(a = 2\), \(b = 19\), \(c = -10\).
2. Find two numbers whose product is \(a \cdot c = 2 \cdot (-10) = -20\) and whose sum is \(b = 19\). The numbers are \(20\) and \(-1\) because \(20 \cdot (-1) = -20\) and \(20 + (-1) = 19\).
3. Rewrite the middle term: \(2k^2 + 20k - k - 10\).
4. Group and factor: \((2k^2 + 20k) + (-k - 10) = 2k(k + 10) - 1(k + 10)\).
5. Factor out the common term: \((2k - 1)(k + 10)\).

So, the factorization is \((2k - 1)(k + 10)\).

Problem 13: \(4p + 12 + 9\)

This simplifies to \(4p + 21\), which cannot be factored further.

Problem 14: \(9q + 6q + 1\)

This simplifies to \(15q + 1\), which cannot be factored further.

Problem 15: \(12l^2 - 17l - 99\)

We need to factorize \(12l^2 - 17l - 99\).

1. Identify coefficients: \(a = 12\), \(b = -17\), \(c = -99\).
2. Find two numbers whose product is \(a \cdot c = 12 \cdot (-99) = -1188\) and whose sum is \(b = -17\). The numbers are \(-44\) and \(27\) because \(-44 \cdot 27 = -1188\) and \(-44 + 27 = -17\).
3. Rewrite the middle term: \(12l^2 - 44l + 27l - 99\).
4. Group and factor: \((12l^2 - 44l) + (27l - 99) = 4l(3l - 11) + 9(3l - 11)\).
5. Factor out the common term: \((4l + 9)(3l - 11)\).

So, the factorization is \((4l + 9)(3l - 11)\).

Problem 16: \(2m^2 + 63m + 145\)

We need to factorize \(2m^2 + 63m + 145\).

1. Identify coefficients: \(a = 2\), \(b = 63\), \(c = 145\).
2. Find two numbers whose product is \(a \cdot c = 2 \cdot 145 = 290\) and whose sum is \(b = 63\). The numbers are \(29\) and \(34\) because \(29 \cdot 34 = 290\) and \(29 + 34 = 63\).
3. Rewrite the middle term: \(2m^2 + 29m + 34m + 145\).
4. Group and factor: \((2m^2 + 29m) + (34m + 145) = m(2m + 29) + 5(2m + 29)\).
5. Factor out the common term: \((m + 5)(2m + 29)\).

So, the factorization is \((m + 5)(2m + 29)\).

Problem 17: \(4n^2 + 4n + 121\)

This quadratic expression does not factorize into real numbers because the discriminant (\(b^2 - 4ac\)) is negative. Therefore, it cannot be factored over the real numbers.

Problem 18: \(15t^2 - 20t + 5\)

We need to factorize \(15t^2 - 20t + 5\).

1. Factor out the greatest common factor: \(5(3t^2 - 4t + 1)\).
2. Factor the quadratic expression inside: \(3t^2 - 4t + 1\).
- Identify coefficients: \(a = 3\), \(b = -4\), \(c = 1\).
- Find two numbers whose product is \(a \cdot c = 3 \cdot 1 = 3\) and whose sum is \(b = -4\). The numbers are \(-3\) and \(-1\) because \(-3 \cdot (-1) = 3\) and \(-3 + (-1) = -4\).
- Rewrite the middle term: \(3t^2 - 3t - t + 1\).
- Group and factor: \((3t^2 - 3t) + (-t + 1) = 3t(t - 1) - 1(t - 1)\).
- Factor out the common term: \((3t - 1)(t - 1)\).

So, the factorization is \(5(3t - 1)(t - 1)\).

Problem 19: \(2x^2 + x + 6\)

This quadratic expression does not factorize into real numbers because the discriminant (\(b^2 - 4ac\)) is negative. Therefore, it cannot be factored over the real numbers.

Problem 20: \(3x^2 + 4x - 1\)

We need to factorize \(3x^2 + 4x - 1\).

1. Identify coefficients: \(a = 3\), \(b = 4\), \(c = -1\).
2. Find two numbers whose product is \(a \cdot c = 3 \cdot (-1) = -3\) and whose sum is \(b = 4\). The numbers are \(3\) and \(-1\) because \(3 \cdot (-1) = -3\) and \(3 + (-1) = 4\).
3. Rewrite the middle term: \(3x^2 + 3x - x - 1\).
4. Group and factor: \((3x^2 + 3x) + (-x - 1) = 3x(x + 1) - 1(x + 1)\).
5. Factor out the common term: \((3x - 1)(x + 1)\).

So, the factorization is \((3x - 1)(x + 1)\).

Final Answer:

\[
\boxed{(3y + 1)(y + 5), (6z + 1)(z - 2), (5w - 2)(w - 1), \text{not factorable}, -5(y + 2), 44, 3(9m + 2), (y - 1)(y - 4), 9(4z + 7), \text{not factorable}, (2d + 1)(4d - 3), (2k - 1)(k + 10), 4p + 21, 15q + 1, (4l + 9)(3l - 11), (m + 5)(2m + 29), \text{not factorable}, 5(3t - 1)(t - 1), \text{not factorable}, (3x - 1)(x + 1)}
\]